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240-2007&2008-2-M20-May2008

240-2007&2008-2-M20-May2008 - MATHEMATICS&...

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Unformatted text preview: MATHEMATICS & COMPUTER - KUWAIT UNIVERSITY SCIENCE DEPARTMENT Differential equations Date: May 10, 2008 Math-240 Each question : 5 marks Second Examination Duration : 90 Minutes Calculators and mobile phones are not allowed _ Answer all questions 1. (a) Show that the functions f1($) r em, f2[a:) = x'2 and f3(r) : 112—2l1133 ( x > 0) are linearly independent functions. (b) Solve the equation, (175+ D4~9D79)y=0. 2. Verify the following (a) (D + e‘“)(e—ID 7 1) 7t (FIB —1)(D + e”). 1 _M _ Ike—“1 “3) - k: 3. If yl : 171” is a solution of the homogeneous differential equation, 45:23:” + Smy' 7 33,! = 0. Find the general solution of the nonhomogeneous differential equation, 4.223,!" + 8mg” — 3y = 4:33. 4. Use the method of variation of parameters to solve the equation, y”+y=ten2m secx. 5. Make use of inverse differential operators formula to solve the equation, [D—2)(D2 + 3D + 2)y:e2“c (3cos(3;1:) + 24x) . GOOD LUCK See formulas in the backside OO 99 DIFFERENTIAL EQUATIONS (MATHZéO) - INVERSE OPERATOR FORMULAS 1. 2. Q)! D 6am: 5. 1 if fugaéo 1 ax“ I e“ _ ‘39 D-a’ce ‘k1a= 1f ¢(a)#0 S'mbzz: _ 1 8'1an 2 2 D2+a2 Cosbx m a2_b2 00353: , a 7&2; 51110.9: Vi -—coscm: . Dz+a2 cosaa: _2a Sinam D (8 10:8 "JD—My OO 99 / Math-240 second mid-term April 30, 2008 w 1-a) £151 + Cgm’i 7 C3:c‘2 lnm = U. a: > D The wronskian of the functions is e“ 55—2 1'2 1111‘ 6’ 3:4 9:41:13: W = e” 722—3 72m‘3 111:5 + 1’3 z 0 723—3 7 :5’2 72x‘3 113:: + :5'3 7 35—2 lnz 31 (ix—4 —5x‘4 + 6.2—4 lnx 0 612—4 7 32—2 7522—4 + 632—41111: 7 55—21113: = 81(4sc’7 + {M‘s +775) # 0 Therefore the functions are linearly independent. (135+ D4—QD—9)y:0 Auxiliary equation is m5 + m4 7 9m 7 9 = D :> m'1 [m+1 )—9(m-.—1) = (7114—9) (m+1) = (mg—3) (m2+3)(m+1) :0 The roots are (7M3, J3, 71, :jfi) The solution is y = cleflfiw -§- 026/51 + 036” + C4 cos + C5 sin 1-1)) 2-21) (B + e’”)(e’ID —1)#(e'wD — 1)(D + 6'”) (D + Fate—“D A1)? = (D + €‘”i(e‘“Dy - y) = Die—’Dy - y) + E‘mDy — e‘fid : e’mDZy 7 6““Dy 7 By + e‘hDy — 3—3519; 2 [ta—“D2 + (Ea—2"“ — 5"“ —1)D — 6"] y and (ti—xD —1)(D + e’fly = (5*!) 71)(Dy + 84y) = e'xDLDy + (fry) — (Dy + 8—5619) = e‘”D2y + e‘ngy — €219: — Dy 7 611,: = [e’a‘D2 + (e’zi - 1) D 7 (9’23 + 6”” 31 So that. [€_$D2 + (ea—2‘” — e"E — 1) D — 6"] # [exl'Jz + (62‘ + 1) y + £21 7 (e’h + 6%)] l mks—“I 2"” = kl (D + it)" ($k8_”) e‘flD" (1“) 2“”.lc! _— “[13: = — (D+a)ke ] k! k! k! 3) If yl : ml-flis one solution of the equation, 45523;” + Smy' — 3y = 0., - - - - (1) and 4:33;” +8my' 7 3y : 4:03., ----(2) To find the general solution let ,y9 = “@1112: y; = 74.1m1/2 +%,Um—1/2’ y; = ENE/2 + vim—U2 _ inn—3m Substitute in (2) 4:152 (WWI/2 + "WI—“2 - ivx’afl) -- 8:6(v'w1f2 + %x_1/2v) — 3UCL'1/2 = 43:3 41*”m5/z +12m3/2v’ : 4:53 :> 2:” + in’ = 11/2 :c 3 wt = ,0” the]: wt __ _w : 1.1/‘2 x —d 2 Ju=ef$ 35:13 so that (33w)’=$7/2 =~ $3w=§$w2+c1 4 Essa/'7 —%C1$_2 + (12 4 . yg = 11$”? : 2311’? (—zafl - film” + Cg) : 7%clz’3/2 + owl/2 + —:n Let u; : 11' Linear in w with integrating factor I 3 Ema/1 3 w=v +clm'::>'v= 45 OO 99 4) (D2 +1)y = tau2 3 seem, u - A (l) (D2+1)y20. ----(2) m2+1:0. 3} m:::i yc=c1cosm+625inm Let yp = A(w)cos:zc +B(J) sinm 7-- y; = A’cosa: 7 Asincz+ B’sinm + Bcosw ----(5) Set A'cosa:+B’sinx:D y; : 7Asinx+Bcosm ----(7) y; = —A’sinw 7Acoszt + B'cosm 735mm - Substitute (4) and (8) into we get -—A’sinm+B’cosx :tan21 secx (9) A’cosx+B’si.um=0 ~~(6) Solve the two equations 3' = tle2 a: and A’ : 7tan3 3 A = [71221423 wdm = f(l 7 seCQmjtanIda: : 71ncosx 7 étanix B:ftan2:cdz:f(sec2$—1)d$ =tanI—m Substitute in [4] My : 7 (ha cosm + i; tan2 cosx +(t3111 — .135in = lésincvtanm 7 cosxlncosa: — 0351113: 99 2 ya + y? 5) (pimp? 7 313 + 2 11:62 (3mm) + 24:!) (1) ) (m—2)(m+2](m+1):0, m; ya = C15 “+021? +035 y” : [62: (3mm + 2422)] 1 =ezr 300$ 3x + 24x yp (D+2—2)((D+2)2+3[D+2)—2)( ( ) ) a: 1 VP : 9' W (“04311 +- 24m) 1 . yp:€21(D2—-L?5+—12)(5m(3$) + 12H) 1 1 7 2m —_ _ —v—_‘ 2 ypwe (D2+7D+12)5m(3$)+(D2+7D+12)12$] 1 1 2:: sin(39:) + 12 m2] yp=6 _(_9+7D+12) m l _ 7 yp = e227 5111(31) + — WE W %D2)12 $2] _ 2; (7D — 3) yp—e —-r—-—(49D2 _g) 9? = 822». [H 511mm) + (1- 1.723 + (1%31132) I2] sin(3::) +— (1 _ 1—723 + $192) :62] (am—3)? —9) E135 (21005(3m) 7 35in(3:c)) + ($2 _ %I + % ] 9'9 = y: + L'p yp : 62: OO 99 ...
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240-2007&2008-2-M20-May2008 - MATHEMATICS&...

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